TITLE: Is this number friends relationship transitive?
QUESTION [6 upvotes]: I am calling two real numbers $a,b$ friends, if there exists a non-constant integer polynomial $p(x)$, such that $p(a)-p(b)=0$. The relation is obviously symmetric and reflexive. Is the relationship also transitive?
( i.e. Can we state for all numbers $a,b,c \in \mathbb R$ that if there is a polynomial $p_1(x) \in \mathbb Z[x]$ and $deg(p_1)>0$ such that $p_1(a)-p_1(b)=0$, and a polynomial $p_2(x) \in \mathbb Z[x]$ and $deg(p_2)>0$ such that $p_2(b)-p_2(c)=0$, does there also exist polynomial $p_3(x) \in \mathbb Z[x]$ and $deg(p_3)>0$ such that $p_3(a)-p_3(c)=0$ )

REPLY [2 votes]: I'm making another answer since my previous answer is so long ago, it would probably go unnoticed if edited.
The answer is no, the relation is not transitive. 
I'll use your observation that any number $a$ is related to $-a-1$ (take $x(-x-1)$). Through $P(x) = x^2$, any number is related to its negative. I'll just show that not any number $a$ is related to $a+1$.
Suppose it is, through $P(x) = \sum_{k=0}^n c_kx^k$. Then
$$
P(a+1)-P(a)=0= \sum_{k=1}^n c_k \sum_{j=0}^{k-1} {k \choose j} a^j =: Q(a) = c_1 + c_2(1 + 2a) + \ldots + c_n \sum_{j=0}^{n-1} {n \choose j} a^j.
$$
Now $Q$ could be the zero polynomial, but then either $P=0$ (not allowed) or $a$ must take one of finitely many (algebraic) values, e.g. $a = -1/2$ allows $c_2$ to be non-zero. In all other cases, $a$ satisfies a non-zero integer polynomial, so must be algebraic. So whenever $a$ is transcendental, $a \sim -a-1 \sim a+1$, but not $a \sim a+1$.